Currently I'm taking a first course on Projective Geometry and I'm working on the following problem :

Given a homogeneous polynomial $F$ of degree $2$ and $n+1$ variables we consider the quadric in $\mathbb P^n(\mathbb R)$ given by: $\mathcal{Q}=\{[v]\in \mathbb{P}^n(\mathbb R): F(v)=0\}$

and the task is to give a projective classification of $\mathcal{Q}$ given that we already know the afine classification.

What I know so far is that the same polynomial $F$ defines a quadric in $\mathbb R^{n+1}$ $$\mathcal{\tilde Q}=\{v\in \mathbb{R}^{n+1}: F(v)=0\},$$ since $w\in[v] \implies w=\lambda v, \lambda\in\mathbb R \setminus \{0\}$ and therefore $F(w)=\lambda^2F(v)=0$. Now, for this quadric $\mathcal{\tilde Q}$ we have its affine classification given by an affine isomorphism that maps it to the"canonial" form.

If I'm not mistaken, a projective transformation is induced by a linear transformation $T:\mathbb R^{n+1} \rightarrow\mathbb R^{n+1}$ by taking: $$\mathbb P(T):\mathbb P^n(\mathbb R) \rightarrow :\mathbb P^n(\mathbb R) \\ [v]\mapsto [T(v)]$$but what I'm not sure and I'd like to know, is how do we go about and define a projective isomorphism from the affine isomorphism?

Any help will be greatly appreciated!


Note that if $F$ is homogeneous, since it comes from a bi-linear form, it has quadratic terms and so all terms must be quadratic. So $F$ is of the form

$$ F(X_1, \dots, X_{n+1}) = \sum_{i = 1}^{n+1}a_iX_i^2 + \sum_{i,j}b_{i,j}X_iX_j. $$

In particular, we have that

$$ \frac{\partial F}{\partial X_k}(X_1,\dots,X_{n+1}) = 2X_k + \sum_jb_{k,j}X_j. $$

and so $0 = F(0) = \nabla F(0)$. This means that the quadric associated with $F$ has a singular point and so it is affinely equivalent to one of the form

$$ A = \sum_{i = 1}^p X_i^2 - \sum_{j = p+1}^rX_j^2 $$

Let $f : x \in \mathbb{R}^{n+1} \mapsto Dx + b \in \mathbb{R}^{n+1}$ be the affine transformation so that $Af = F$ and in particular, one quadric is mapped to the other. Since $A$ and $F$ are homogenous, they are of the form

$$ A = x^tQx, \quad F = x^tBx $$

and so since $Af = F$ we have that, as polynomials, their cuadratic terms must coincide. Factoring the composite of $A$ and $f$, then, we get

$$ A = x^tQx = x^tD^tBDx = (Dx)^tBDx = F(Dx) $$

and thus we can keep the linear part of the isomorphism, which will still behave as desired.


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